假设已经有了前序遍历和中序遍历的结果，通过一个算法重建这棵树

rt,

分析与解法

前序: a b c d e f

后序: d b a e c f

“a”是前序遍历节点的第一个元素,它把中序遍历的结果分为“db”和“ecf”两个部分，这两部分也是“a”的左右子树的遍历结果。

如果能够找到前序遍历中对应的左子树和右子树，就可以把“a”作为当前的根节点，然后依次递归下去，这样就能够依次恢复左子树和右子树的遍历结果。

C# Codes

using System;
using System.Collections.Generic;
using System.Text;

namespace ConsoleApp
{
class Program
{
static void Main(string[] args)
{
TreeNode<int> node1 = new TreeNode<int>();
node1.Data = 1;

TreeNode<int> node2 = new TreeNode<int>();
node2.Data = 2;

TreeNode<int> node3 = new TreeNode<int>();
node3.Data = 3;

TreeNode<int> node4 = new TreeNode<int>();
node4.Data = 4;

TreeNode<int> node5 = new TreeNode<int>();
node5.Data = 5;

TreeNode<int> node6 = new TreeNode<int>();
node6.Data = 6;

TreeNode<int> node7 = new TreeNode<int>();
node7.Data = 7;

TreeNode<int> node8 = new TreeNode<int>();
node8.Data = 8;

TreeNode<int> node9 = new TreeNode<int>();
node9.Data = 9;

TreeNode<int>[] nodes1 ={ node1, node2, node4, node7, node3, node5, node8, node6, node9 };
TreeNode<int>[] nodes2 ={ node7, node4, node2, node1, node8, node5, node3, node6, node9 };

TreeNode<int> root = ReBuild<int>.ReBuildTree(nodes1, nodes2, 0, 0, nodes1.Length);

Tree<int> tree2 = new Tree<int>();
tree2.Root = root;

Tree<int>.Print2(tree2);

Console.ReadKey();
}
}

class Tree<T>
{
public TreeNode<T> Root;

/// <summary>
/// 广度优先遍历二叉树
/// </summary>
public static void Print2(Tree<T> tree)
{
if (tree.Root == null)
{
return;
}
else
{
Queue<TreeNode<T>> queue = new Queue<TreeNode<T>>();
queue.Enqueue(tree.Root);

while (queue.Count != 0)
{
int currentLevelCount = queue.Count;

for (int i = 0; i < currentLevelCount; i++)
{
TreeNode<T> node = queue.Dequeue();

if (node.Data != null)
{
Console.Write(node.Data);
}

if (node.LeftSon != null)
{
queue.Enqueue(node.LeftSon);
}

if (node.RightSon != null)
{
queue.Enqueue(node.RightSon);
}
}

Console.WriteLine();
}
}
}
}

class TreeNode<T>
{
public T Data;

public TreeNode<T> LeftSon;
public TreeNode<T> RightSon;
}

class ReBuild<T>
{
public static TreeNode<T> ReBuildTree(TreeNode<T>[] preOrder, TreeNode<T>[] inOrder, int preIndex, int inIndex, int length)
{
if (preOrder == null || length == 0)
{
return null;
}
else if (preOrder.Length != inOrder.Length)
{
return null;
}
else if (length == 1)
{
return preOrder[preIndex];
}

for (int i = inIndex; i < inIndex + length; i++)
{
if (preOrder[preIndex] == inOrder[i])
{
preOrder[preIndex].LeftSon = ReBuildTree(preOrder, inOrder, preIndex + 1, inIndex, i - inIndex);
preOrder[preIndex].RightSon = ReBuildTree(preOrder, inOrder, preIndex + i - inIndex + 1, i + 1, length - i + inIndex - 1);
}
}

return preOrder[preIndex];
}
}
}